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a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{15.16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{\left(x-1\right)\times x}=\dfrac{15}{16}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x-1}-\dfrac{1}{x}=\dfrac{15}{16}\)
\(1-\dfrac{1}{x}=\dfrac{15}{16}\)
\(\dfrac{1}{x}=1-\dfrac{15}{16}=\dfrac{16}{16}-\dfrac{15}{16}\)
\(\dfrac{1}{x}=\dfrac{1}{16}\)
\(\Rightarrow x=16\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{14.15}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{14}-\frac{1}{15}\)
\(A=1-\frac{1}{15}\)
\(A=\frac{14}{ }15\)
khó quá sai thì thôi nha
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/49 - 1/50
= 1/1 - 1/50
= 49/50
1/1.2 + 1/2.3 + 1/3.4 +.....+ 1/49.50
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/49 - 1/50
= 1 - 1/50
= 49/50
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{999}-\frac{1}{1000}+1\)
\(\frac{1}{1}-\frac{1}{1000}+1\)
\(\frac{999}{1000}+1\)
\(\frac{1999}{1000}\)
a) \(1010+1111+1212+.....+9898+9999\)\(=\frac{\left(1010+9999\right)\cdot\left(\frac{9999-1010}{1111-1010}+1\right)}{2}\)\(=\frac{11009\cdot\left(\frac{8989}{101}+1\right)}{2}\)\(=\frac{11009\cdot\left(89+1\right)}{2}\)\(=\frac{11009\cdot90}{2}\)\(=\frac{990810}{2}\)\(=495405\)
a) Khoảng cách của dãy số là:
1111-1010=101;1212-1111=101;...
Số số hạng của dãy số là:
(9999-1010):101+1=90(số)
Tổng:
(1010+9999)*90:2=495405
Đ/s:495405
Nhớ k mk nha!
Đề bài
=1-1/2+1/2-1/3+.....+1/15-1/16
=1-1/16
=15/16
15/16 nha